Another way to find coordinates of two remain vertices knowing given two vertices?

Question

Let be given the polyhedron ABCD with $AB = 4 \sqrt{3}$, $AC=1$, $BC=\sqrt{41}$, $AD=\sqrt{17}$, $CD=4$, and $BD=5$. I am trying to find coordinates of two vertices $B$ and $D$ knowing $A(0,0,0)$ and $C(1,0,0)$. I tried

Clear[a, b, c];
a = {0, 0};
c = {1, 0};
b = {x, y};
Solve[{EuclideanDistance[a, b] == 4 Sqrt[3], EuclideanDistance[c, b] == Sqrt[41]}, {x, y}]

I got

{{x -> 4, y -> -4 Sqrt[2]}, {x -> 4, y -> 4 Sqrt[2]}}

And then, I solved

Clear[a, b, c]
a = {0, 0, 0};
c = {1, 0, 0};
b = {4, 4 Sqrt[2], 0};
d = {x, y, z};
d /. Solve[{EuclideanDistance[a, d] == Sqrt[17], 
   EuclideanDistance[b, d] == 5, EuclideanDistance[c, d] == 4}, {x, y,
    z}, Reals]

{{1, 2 Sqrt[2], -2 Sqrt[2]}, {1, 2 Sqrt[2], 2 Sqrt[2]}}

How can I get coordinates of two points B and D with another way?

Answer 1

To make things easier, you can assume that the first 3 points a,b,c are in the x-y plane plane. This will give you some solutions. All other solutions are obtained by rotating this solutions around the a-c axis (here the x-axis). With this assumptions:

Clear["Globals`*"]

a = {0, 0, 0};
c = {1, 0, 0};
b = {b1, b2, 0};
d = {d1, d2, d3};
sol = {b1, b2, d1, d2, d3} /. 
  Solve[{EuclideanDistance[a, b] == 4 Sqrt[3], 
    EuclideanDistance[b, c] == Sqrt[41], 
    EuclideanDistance[a, d] == Sqrt[17], EuclideanDistance[c, d] == 4,
     EuclideanDistance[b, d] == 5}, Join[b[[;; 2]], d], Reals]

This gives 4 different solutions. We can draw these simplex by:

Graphics3D[{Simplex[{a, b, c, d}] /. #}, Axes -> True, 
   AxesLabel -> {"x", "y", "z"}] & /@ rules

All other solutions are rotations of the above around the x axis.

Answer 2

I'm not sure how to get mathematica to do this elegantly.

Both $B$ and $D$ must lie on particular circles. Namely B={4,4Sqrt[2]Cos@θ,4Sqrt[2]Sin@θ} and D={1,4Cos@ϕ,4Sin@ϕ}. Knowing this, we can do

Solve[Norm[{4,4Sqrt[2]Cos@θ,4Sqrt[2]Sin@θ}-{1,4Cos@ϕ,4Sin@ϕ}]==5,{θ,ϕ},Reals]

to find ϕ=θ±π/4. So there are infinitely many such $A,B,C,D$.

Here's a visualization of half of the infinitely many solutions (the other half have $D$ 'lagging' $B$ instead of 'leading').